Question #4cb9a

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Question #4cb9a

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Answer 1 · 2016-10-26T07:45:27

Let the length of each side of an equilateral triangle be a.Its any altitude, the perpendicular from any vertex to opposite side will bisect the base. If length of the altitude is h then applying pythagoras theorm we can write

$a^{2} = h^{2} + {(\frac{a}{2})}^{2}$ $a^2=h^2+(a/2)^2$

$\Rightarrow h^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3 a^{2}}{4}$ $=>h^2=a^2-a^2/4=(3a^2)/4$

$\Rightarrow h = \frac{\sqrt{3} a}{2}$ $=>h=(sqrt3a)/2$

In our problem $h = 12 \sqrt{3}$ $h=12sqrt3$

So $12 \sqrt{3} = \frac{\sqrt{3}}{2} a$ $12sqrt3=sqrt3/2a$
$\Rightarrow a = 24$ $=>a=24$

So length of each side of the equilateral triangle is $a = 24$ $a= 24$

Hence the perimeter of the triangle will be $= 3 a = 3 \cdot 24 = 72$ $=3a=3*24=72$

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